Wave Optics

JEE Main level, single correct, medium

39 of 191

Question 39 of 191

In a double slit experiment the distance between the slits is 0.1 cm and the screen is placed at 50 cm from the slits plane. When one slit is covered with a transparent sheet having thickness tt and refractive index n(=1.5)n(=1.5), the central fringe shifts by 0.2 cm. The value of tt is ____\_\_\_\_ cm.

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Answer: (C) 8×10−48 \times 10^{-4}

Given: d=0.1d = 0.1 cm, D=50D = 50 cm, n=1.5n = 1.5, Shift =0.2= 0.2 cm
Optical path difference by sheet: Δ=(n−1)t\Delta = (n-1)t
Shift of central fringe: Shift =(n−1)t×Dd= \frac{(n-1)t \times D}{d}
0.2=(1.5−1)×t×500.1=0.5×50×t0.1=250t0.2 = \frac{(1.5-1) \times t \times 50}{0.1} = \frac{0.5 \times 50 \times t}{0.1} = 250t
t=0.2250=8×10−4t = \frac{0.2}{250} = 8 \times 10^{-4} cm

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